1029. Two City Scheduling

Problem

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCostᵢ, bCostᵢ], the cost of flying the iᵗʰ person to city a is aCostᵢ, and the cost of flying the iᵗʰ person to city b is bCostᵢ.

Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.

https://leetcode.com/problems/two-city-scheduling/

Example 1:

Input: costs = [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation:
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859

Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086

Constraints:

• 2 * n == costs.length
• 2 <= costs.length <= 100
• costs.length is even.
• 1 <= aCostᵢ, bCostᵢ <= 1000

Test Cases

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class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:

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import pytest

from solution import Solution


@pytest.mark.parametrize('costs, expected', [
    ([[10,20],[30,200],[400,50],[30,20]], 110),
    ([[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]], 1859),
    ([[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]], 3086),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, costs, expected):
    assert sol.twoCitySchedCost(costs.copy()) == expected

Thoughts

不妨先让所有人都去城市 A，总成本为 ΣaCostᵢ。然后需要把 n 个人改派到城市 B。如果把第 i 个人改派到城市 B，成本的变化为 bCostᵢ - aCostᵢ，显然这个值越小越好。

所以对 costs 按照 bCostᵢ - aCostᵢ 排序，前边 n 个人去城市 B，后边 n 个人去城市 A，这样总的成本最低。

时间复杂度 O(n log n)。用 in-place 排序，附加的空间复杂度 O(1)。

Code

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class Solution:
    def twoCitySchedCost(self, costs: list[list[int]]) -> int:
        costs.sort(key=lambda c: c[1] - c[0])
        n = len(costs) >> 1
        return sum(costs[i][1] for i in range(n)) + sum(costs[i][0] for i in range(n, n<<1))

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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